[PAT A1008]Elevator
题目描述
1008 Elevator （20 分)The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.
For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.
输入格式
Each input file contains one test case. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100.
输出格式
For each test case, print the total time on a single line.
输入样例
3 2 3 1
输出样例
41
解析
题目很简单，大致意思就是输入n个数，每个数表示电梯将要在哪一层停靠，每上升一层需要6秒钟，下降一层需要4秒钟，停靠一层开关门时间需要5秒钟，要我们计算总的用时是多少。这个题目就要要求一遍能通过是最好的。
#include<iostream>
using namespace std;
int main()
{
int n;scanf("%d", &n);int pos = 0, time = 0;//电梯当前所在位置for (int i = 0; i < n; i++) {
int des; //目标楼层scanf("%d", &des);if (des > pos) {
time += (des - pos) * 6;pos = des;}else {
time += (pos - des) * 4;pos = des;}}time += n * 5;printf("%d", time);return 0;
}